Question: Solve for $x$ : $6\sqrt{x} - 3 = 3\sqrt{x} + 2$
Explanation: Subtract $3\sqrt{x}$ from both sides: $(6\sqrt{x} - 3) - 3\sqrt{x} = (3\sqrt{x} + 2) - 3\sqrt{x}$ $3\sqrt{x} - 3 = 2$ Add $3$ to both sides: $(3\sqrt{x} - 3) + 3 = 2 + 3$ $3\sqrt{x} = 5$ Divide both sides by $3$ $\frac{3\sqrt{x}}{3} = \frac{5}{3}$ Simplify. $\sqrt{x} = \dfrac{5}{3}$ Square both sides. $\sqrt{x} \cdot \sqrt{x} = \dfrac{5}{3} \cdot \dfrac{5}{3}$ $x = \dfrac{25}{9}$